/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 (a) A process called gaseous dif... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(a) A process called gaseous diffusion is often used to separate isotopes of uranium-that is, atoms of the elements that have different inasses, such as \(^{235} \mathrm{U}\) and \(^{238} \mathrm{U}\) . The only gaseous compound of uranium at ordinary temperatures is uranium hexafluoride, \(\mathrm{UF}_{6}\) . Speculate on how 235 \(\mathrm{UF}_{6}\) and \(^{238} \mathrm{UF}_{6}\) inolecules might be separated by diffusion. (b) The molar masses for \(^{235} \mathrm{UF}_{6}\) and \(^{238} \mathrm{UF}_{6}\) molecules are 0.349 \(\mathrm{kg} / \mathrm{mol}\) and \(0.352 \mathrm{kg} / \mathrm{nol},\) respectively. If uranium hexafluoride acts as an ideal gas, what is the ratio of the root-mean- square speed of \(^{235} \mathrm{UF}_{6}\) molecules to that of \(^{238} \mathrm{UF}_{6}\) molecules if the temperature is uniform?

Short Answer

Expert verified
The root-mean-square speed of \( ^{235}UF_6 \) is approximately 1.0043 times that of \( ^{238}UF_6 \).

Step by step solution

01

Understanding Gaseous Diffusion

Gaseous diffusion is a method employed to separate isotopes of the same element based on their mass. In this process, lighter molecules (those with less mass) move through a barrier or medium more quickly than heavier ones. Since the molecules of interest have different molar masses, gaseous diffusion can be used to separate them by exploiting their different speeds during the diffusion process.
02

Expressing the Root-Mean-Square Speed Formula

The root-mean-square (rms) speed of a gas is related to its molar mass and the temperature of the system. The formula is given by:\[v_{rms} = \sqrt{\frac{3kT}{m}} = \sqrt{\frac{3RT}{M}}\]where \(v_{rms}\) is the root-mean-square speed, \(k\) is the Boltzmann constant, \(T\) is the temperature, \(m\) is the mass of one molecule, \(R\) is the ideal gas constant, and \(M\) is the molar mass of the gas. For a given temperature, the speed is inversely proportional to the square root of the molar mass.
03

Calculating the Speed Ratio

To find the ratio of the root-mean-square speed of \(^{235}UF_6\) to \(^{238}UF_6\), we use the formula derived from the previous step and consider that both gases are at the same temperature:\[\frac{v_{rms, ^{235}UF_6}}{v_{rms, ^{238}UF_6}} = \sqrt{\frac{M_{^{238}UF_6}}{M_{^{235}UF_6}}}\]Substitute the given molar masses: \(0.352 \text{ kg/mol} \text{ for } ^{238}UF_6\) and \(0.349 \text{ kg/mol} \text{ for } ^{235}UF_6\), resulting in:\[\frac{v_{rms, ^{235}UF_6}}{v_{rms, ^{238}UF_6}} = \sqrt{\frac{0.352}{0.349}}\]
04

Simplifying and Solving the Ratio

Compute the ratio:\[\frac{v_{rms, ^{235}UF_6}}{v_{rms, ^{238}UF_6}} = \sqrt{1.0086} \approx 1.0043\]This shows that the \( v_{rms} \) of \( ^{235}UF_6 \) is approximately 1.0043 times the \( v_{rms} \) of \( ^{238}UF_6 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Isotope Separation
Isotope separation is a crucial process in the nuclear industry, especially when dealing with uranium. Isotopes are atoms of the same element but with different masses due to varying numbers of neutrons in the nucleus. Uranium has two primary isotopes: - Uranium-235 - Uranium-238 The separation of these isotopes is vital because uranium-235 is used in nuclear reactors and weapons due to its fissile properties, meaning it can sustain a nuclear chain reaction.
Gaseous diffusion is one common method of separating these isotopes. This technique leverages the slight difference in masses of the isotopes' molecules. As gases, lighter molecules such as - uranium-235 hexafluoride ( ^{235}UF_6) migrate more rapidly through a barrier compared to the heavier uranium-238 hexafluoride ( ^{238}UF_6). This difference in movement speed is the basis for their separation.
Uranium Hexafluoride
Uranium hexafluoride ( UF_6) is a compound that plays a pivotal role in the enrichment of uranium isotopes. It is the only uranium compound that exists as a gas at relatively low temperatures, making it ideal for gaseous diffusion.
Uranium hexafluoride is a combination of one uranium atom and six fluorine atoms. Its significance arises from:
  • Ease of conversion from solid to gas
  • Its reactive nature, which allows for efficient processing
  • Stability at the temperatures where diffusion occurs
To handle and process uranium, it is transformed into UF_6. Once in gaseous form, the slight mass differences between molecules containing uranium-235 and uranium-238 enable their separation during diffusion processes.
Root-Mean-Square Speed
Root-mean-square (rms) speed is a concept frequently used to describe the speed of gas particles in a system. It helps in understanding how different factors affect the movement of molecules. The formula for calculating rms speed is:\[v_{rms} = \sqrt{\frac{3RT}{M}}\]Here, \(v_{rms}\) is the root-mean-square speed, \(R\) is the gas constant, \(T\) is the temperature, and \(M\) is the molar mass.
From the equation, we can see:
  • The rms speed is inversely proportional to the square root of the molar mass. This means that lighter molecules move faster.
  • Temperature directly affects the rms speed; higher temperatures lead to higher speeds.
In isotope separation, because uranium-235 UF_6 is slightly lighter than uranium-238 UF_6, the rms speed is higher for the former, facilitating their separation.
Ideal Gas Law
The ideal gas law is an essential principle in understanding gas behavior in variable conditions. It is mathematically expressed as:\[PV = nRT\]where:
  • \(P\) is the pressure
  • \(V\) is the volume
  • \(n\) is the number of moles
  • \(R\) is the ideal gas constant
  • \(T\) is the temperature
This equation helps in predicting how gases will react under different conditions. It assumes no interactions between molecules other than elastic collisions and that gases occupy no volume of their own. This makes it a helpful approximative tool in gaseous diffusion processes like isotope separation.
The ideal gas law lays the foundation for further understanding gas properties, which include diffusion and effusion. These properties are leveraged in the gaseous diffusion method, where lighter gases at a given temperature will diffuse faster, helping isolate specific isotopes such as uranium-235 in uranium hexafluoride.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A metal tank with volume 3.10 L will burst if the absolute pressure of the gas it contains exceeds 100 atm. (a) If 11.0 mol of an ideal gas is put into the tank at a temperature of \(23.0^{\circ} \mathrm{C},\) to what temperature can the gas be warmed before the tank ruptures? You can ignore the thermal expansion of the tank. (b) Based on your answer to part (a), is it reasonable to ignore the thermal expansion of the tank? Explain.

A flask with a volume of 1.50 \(\mathrm{L}\) , provided with a stopock, contains ethane gas \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) at 300 \(\mathrm{K}\) and atmospheric pressure \(\left(1.013 \times 10^{5} \mathrm{Pa}\right) .\) The molar mass of ethane is 30.1 \(\mathrm{g} / \mathrm{mol}\) . The system is warmed to a temperature of \(380 \mathrm{K},\) with the stopcock open to the atmosphere. The stopcock is then closed, and the flask is cooled to its original temperature. (a) What is the final pressure of the ethane in the flask? (b) How many grams of ethane remain in the flask?

An empty cylindrical canister 1.50 \(\mathrm{m}\) long and 90.0 \(\mathrm{cm}\) in diameter is to be filled with pure oxygen at \(22.0^{\circ} \mathrm{C}\) to sture in a space station. To hold as thuch gas as possible, the absolute pressure of the oxygen will be 21.0 atm. The molar mass of oxygen is 32.0 \(\mathrm{g} / \mathrm{mol}\) . (a) How many tholes of oxygen does this canister hold? (b) For someone lifting this canister, by how many kilo-grams does this gas increase the mass to be lifted?

A Jaguar XK8 convertible has an eight-cylinder engine. At the beginning of its compression stroke, one of the cylinders contains 499 \(\mathrm{cm}^{3}\) of air at atmospheric pressure \(\left(1.01 \times 10^{5} \mathrm{Pa}\right)\) and a temperature of \(27.0^{\circ} \mathrm{C}\) . At the end of the stroke, the air has been compressed to a volume of 46.2 \(\mathrm{cm}^{3}\) and the gauge pressure has increased to \(2.72 \times 10^{5}\) Pa. Compute the final temperature.

The conditions of standard temperature and pressure (STP) are a temperature of \(0.00^{\circ} \mathrm{C}\) and a pressure of 1.00 \(\mathrm{atm}\) . (a) How many liters does 1.00 \(\mathrm{mol}\) of any ideal gas occupy at STP? (b) For a scientist on Venus, an absolute pressure of 1 Venusian-atmosphere is 92 Earth- atmospheres. Of course she would use the Venusian-atmosphere to define STP. Assuming she kept the same temperature, how many liters would 1 mole of ideal gas occupy on Venus?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.